Question: Multiply the following complex numbers: $({3-5i}) \cdot ({-3+2i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({3-5i}) \cdot ({-3+2i}) = $ $ ({3} \cdot {-3}) + ({3} \cdot {2}i) + ({-5}i \cdot {-3}) + ({-5}i \cdot {2}i) $ Then simplify the terms: $ (-9) + (6i) + (15i) + (-10 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -9 + (6 + 15)i - 10i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -9 + (6 + 15)i - (-10) $ The result is simplified: $ (-9 + 10) + (21i) = 1+21i $